How Does Rotating a Polaroid Affect Light Intensity?

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Rotating a Polaroid sheet by 30 degrees results in a light intensity drop calculated using Malus' Law, which states that the intensity after rotation is equal to the initial intensity multiplied by the cosine squared of the angle. The correct equation is I = I_0 cos^2(30), leading to a factor of 0.75 for the intensity reduction. The discussion clarifies that the initial confusion stemmed from misapplying the tangent function instead of the cosine function. Additionally, it emphasizes that power and intensity are related to the electric field strength squared. Understanding these relationships is crucial for accurately determining light intensity changes due to rotation.
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Two sheets of polaroid are oriented so that there is a maximum transmission of light. One sheet is now rotated by 30 degrees, by what factor does the light intensity drop?



OK, the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.


the answer is 0.75 I just can't seem to get there.
Any help would be GREATLY appreciated.
 
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fionamb83 said:
the only equation I could think to use is tanB = N1/N2 but it doesn't seem to work.

Define the terms in this equation: what exactly are B, N1, N2, and how do they relate to the problem you've quoted?
 
Got it!

I feel silly! I realize equation I was trying to use was wrong. I solved it using Malus' Law

intensity after = Intensity before*cos^30

Thanks for the quick reply tho.
 
Malus' Law is spot on. The intensity is greatest when angle=0deg, zero when angle=90 deg. The cos(angle) function fits the bill, especially as the situation is a rotation.
 
Sorry, in the above post I should have said amplitude not intensity.
fionamb83 said:
... Malus' Law

intensity after = Intensity before*cos^30

In the above did you mean I = I_0 \sin 30^o or I = I_0 \sin^2 30^o ?
 
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<br /> I = I_0 \cos^2 30^o<br />

From Cutnell, Physics
 
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fionamb83 said:
<br /> I = I_0 \cos^2 30^o<br />

From Cutnell, Physics
E=E_0 cos(30) where E is the field in Volts/m.

Power (intensity) is proportional to E^2, which in turn is proportional to I^2 (current squared).

\frac{I}{I_0}=\frac{E^2}{E_0^2}=\frac{(E_0 cos(30))^2}{E_0^2}=cos^2(30)=0.75

[n.b. "I" for intensity is not current - I hope that isn't too confusing]

Regards,

Bill
 
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